Can a resistor of known value be used to induce a predictable current, in accordance with Ohm's law?

CONTEXT

I have a 100A shunt and an opamp, which I'm going to use for measuring current through a battery bank. But in order to calibrate it, I need to know the actual current flowing through the circuit. I don't have a 100A (or even 50A) capable meter. I could use my 10A-capable multimeter, but I wouldn't be able to measure at the higher current levels.

METHOD

Can I place a known resistor over a known voltage, and then use Ohm's law to reliably calculate the current? simulate this circuit – Schematic created using CircuitLab

E.g. if I measure the resistance of a dummy load with a multimeter, and then measure the voltage drop over the resistor, can I then reliably calculate the current?

Also, since my cheap voltmeter can measure up to 10A, I'm thinking to test it with an adjustable power supply, by gradually increasing the voltage until the current reaches maybe 5.00A, and then measure the voltage with a separate voltmeter (so that the circuit remains constant), and then use this to calculate the resistance more accurately.

Is this a stupid way to approach it?

• Is there any reason you can't measure 5.00A across the high-powered shunt? Even Chinesium multimeters are pretty accurate nowdays. – Bryan Boettcher Jan 4 '17 at 18:59
• Well that's exactly what I'm proposing, as an initial measurement. But I'm not sure that the current shunt or opamp will give a linear reading for higher currents. So in order to verify that, I was thinking of this approach. – user95482301 Jan 4 '17 at 19:01
• A shunt will be linear. They are fixed-value resistors that behave in a predictable manner when inside of their operating range (mostly, heat). If you go outside that range, the resistance becomes less predictable. – Bryan Boettcher Jan 4 '17 at 19:03
• Is that shunt really 0.75mOhm, or 750uOhm? That's small. – Vladimir Cravero Jan 4 '17 at 19:21
• What do you mean by shunt in this case? A shunt usually is in parallel with a voltmeter, which is calibrated with a proper current scale. – Mike Waters Jan 4 '17 at 19:32

If you have a 100A shunt it is already calibrated (probably within 0.25% or 0.5%). Just disconnect the shunt, apply 0mV to your readout circuit and make it read zero, then apply 75.00mV and make it read 100.0A (repeat if necessary).

There is no need for a high current source unless you doubt the provenance of the shunt.

Either of those methods is a reasonable way to approach it. A clamp on current meter may also be a good option.

If you decide to use your multimeters current function, you'll want to get close to 10 amps to get the best resolution. 9 amps will give you better results than 5. Keep in mind that at 100 amps, resistive heating of your current shunt may throw off the calibration unless you compensate for it, or actively cool the shunt.

One issue you may have with the shunt is that at 100 amps, even a 0.1v signal will result in 10W of power disipation from the shunt. A 1/4W resistor will quickly burn up, and won't be accurate anyway due to the heat. You'll need to go for an even smaller signal, use a very burly resistor, or run the system for only a short burst (like 1 second). Calibrating at a lower voltage using the current function is probably an easier option.

• You can only use a clamp-on meter for AC. Forget using a multimeter's current function. – Mike Waters Jan 4 '17 at 19:24
• There are clamp on meters for DC, they use the hall effect. They are not cheap though en.wikipedia.org/wiki/Current_clamp – Christian Feb 16 '17 at 13:33
• They're not very accurate either, at least in the context of a calibration reference. – pericynthion Jun 23 '17 at 15:54

Op-amp and additional resistors aside for now...

Measuring the current in that circuit can be done with ONLY a voltmeter, provided that you know the value of at least one of those resistors.

Since current in any series circuit is the same, simply measure the voltage drop across a resistor, and then divide that by the resistor's value. (I=E/R)

Ohm's law

• Right. I know Ohm's law, as stated in the question. But I'm concerned about whether it holds up in reality. For example I know I need to make sure that the temperature remains constant for the reading to be reliable. – user95482301 Jan 4 '17 at 19:51
• @user95482301 The datasheet for the shunt resistor should have a temperature coefficient for the resistor. That would allow you to estimate the error due to temperature which, once you've measured any other current value, will be the largest remaining source of measurement error. – Samuel Jan 4 '17 at 20:36

Yes, measuring the voltage across a known resistance is a valid way to measure current. It's done all the time.

However, if you already have a 500 milliohm dummy load, you don't need to also add this shunt. Just measure the voltage across the dummy load. The shunt adds no value. Besides being simpler, the voltage across the dummy load will be much greater, and thus easier to measure without being concerned with noise, etc.

• I think the thing that bothered me, was that Ohm's law seemed too simple. For example, the formula could be expanded to include the temperature coefficient of the material. And the question is basically 'what else do I have to account for'. I now understand that a shunt is made from a material which has a relatively constant resistance across temperature changes. – user95482301 May 24 '17 at 15:03
• The concept of a defacto std 75mV shunt is that it is small enuf to not affect load but big enuf to sense and the power is P=VI or 75mV*100A=7.5W so that temp rise is low the bar has to be big. and R=0.750 mOhm. Also conductors except constantin have a large PTC. Sometimes people choose 100mV shunts more or less, but 75mV is std. – Tony Stewart Sunnyskyguy EE75 May 24 '17 at 18:14